Find the lenght of each side of a rhombus whose diagonals are 24cm and 10cm. long.
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Let ABCD be the given rhombus whose diagonals intersect at O. Then AC = 24cm and BD = 10cm
We know that the diagonals of a rhombus bisect each other at right angles. $\text{OA}=\frac{1}{2}\text{AC}=12\text{cm}$ $\text{OB}=\frac{1}{2}\text{BD}=5\text{cm}$ and $\angle\text{AOB}=90^\circ$ Form right $\triangle\text{AOB},$ we have $\text{AB}^2=\text{OA}^2+\text{OB}^2$ $\Rightarrow\text{AB}^2=\Big[(12)^2+(5)^2\Big]\text{cm}^2$ $\Rightarrow(144+25)\text{cm}^2=169\text{cm}^2$ $\Rightarrow\text{AB}=\sqrt{169}\text{cm}=13\text{cm}$ Hence, each side of a rhombus 13cm.
We know that the diagonals of a rhombus bisect each other at right angles. $\text{OA}=\frac{1}{2}\text{AC}=12\text{cm}$ $\text{OB}=\frac{1}{2}\text{BD}=5\text{cm}$ and $\angle\text{AOB}=90^\circ$ Form right $\triangle\text{AOB},$ we have $\text{AB}^2=\text{OA}^2+\text{OB}^2$ $\Rightarrow\text{AB}^2=\Big[(12)^2+(5)^2\Big]\text{cm}^2$ $\Rightarrow(144+25)\text{cm}^2=169\text{cm}^2$ $\Rightarrow\text{AB}=\sqrt{169}\text{cm}=13\text{cm}$ Hence, each side of a rhombus 13cm.
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